Answer:
³⁸₂₀Ca.
Explanation:
³⁸₁₉K –> __ + ⁰₋₁β
Let ʸₓA represent the unknown.
Thus the equation above can be written as:
³⁸₁₉K –> ʸₓA + ⁰₋₁β
Thus, we can obtain the value of y an x as follow:
38 = y + 0
y = 38
19 = x + (–1)
19 = x – 1
Collect like terms
19 + 1 = x
x = 20
Thus,
ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca
Therefore, the equation is:
³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β
Answer:
6.022 x 10²³; it is a conversion factor between moles and number of particles
Explanation:
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole of hydrogen = 6.022 × 10²³ atoms of hydrogen
238 g of uranium = 1 mole of uranium = 6.022 × 10²³ atoms of uranium
By taking ions:
62 g of NO⁻₃ = 1 mole of NO⁻₃ = 6.022 × 10²³ ions of NO⁻₃
96 g of SO₄²⁻ = 1 mole of SO₄²⁻ = 6.022 × 10²³ ions of SO₄²⁻
Answer:
ΔG°rxn = -69.0 kJ
Explanation:
Let's consider the following thermochemical equation.
N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.
3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ
1) Write the balaced chemical equation:
H2 + 2O2 → 2H2O
2) Infere the molar ratios:
1 mol H2 : 2 mol of water
3) Make the calculus as the direct proportion relation:
[2 mol H2O] / [1 mol H2] * 7 mol H2 = 14 mol H2
As you see you produce the double number of moles of H2O than number of moles of H2 used.
Answer: 14 moles
